reflection and transmission at a potential step · reflection and transmission at a potential step...
TRANSCRIPT
Reflection and Transmission at a Potential Step
Outline - Review: Particle in a 1-D Box - Reflection and Transmission - Potential Step - Reflection from a Potential Barrier - Introduction to Barrier Penetration (Tunneling)
Reading and Applets: .Text on Quantum Mechanics by French and Taylor .Tutorial 10 – Quantum Mechanics in 1-D Potentials .applets at http://phet.colorado.edu/en/get-phet/one-at-a-time
1
Schrodinger: A Wave Equation for Electrons
∂Eψ = �ωψ = −j�
(free-particle)
(free-particle)
..The Free-Particle Schrodinger Wave Equation !
∂�xψ =ψ p kψ = j�
∂tψ
∂x
p2E =
2m
−j�∂
∂tψ = − �
2
2m
∂2ψ
∂x2
Erwin Schrödinger (1887–1961) Image in the Public Domain
2
Schrodinger Equation and Energy Conservation
The Schrodinger Wave Equation
The quantity | |2 dx is interpreted as the probability that the particle can be found at a particular point x (within interval dx)
0 x L n=3
P (x) = |ψ|2dx
2�
Eψ(x) = −
|ψ|2
2m
∂2ψ(x)+ V (x)ψ(x)
∂x2
© Dr. Akira Tonomura, Hitachi, Ltd., Japan. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.
3
Schrodinger Equation and Particle in a Box
EIGENENERGIES for EIGENSTATES for PROBABILITY 1-D BOX 1-D BOX DENSITIES
2�
Eψ(x) = − ∂2ψ(x)
2m+ V (x)ψ(x)
∂x2
ψ(x) =
√2 nπsin
L
( x 2P
L
)(x) = |ψ(x)| dx
ψ3
ψ2
ψ1
0 0L L
x
x
x
x
x
x
4
Semiconductor Nanoparticles Semiconductor Nanoparticles
Red: bigger dots! Blue: smaller dots!
(aka: Quantum Dots) Determining QD energy
using the Schrödinger Equation
2� k2
E1 = 1
2m=
�2π2
2mL2E1 = n2E1
Core Shell
Core Quantum Dot
LRED QD
9EB1
4EB1
EB1
9ER1
4ER1
ER1
LBLUE QD
Photo by J. Halpert Courtesy of M. Bawendi Group, Chemistry, MIT
5
Solutions to Schrodinger s Equation
The kinetic energy of the electron is related to the curvature of the wavefunction
Tighter confinement
Nodes in wavefunction
Higher energy
Higher energy
The n-th wavefunction (eigenstate) has (n-1) zero-crossings
2�−
Even the lowest energy bound state requires some wavefunction curvature (kinetic energy) to satisfy boundary conditions
∂2ψ
2m= (E
∂x2− V (x))ψ
6
• corresponds to a physically meaningful quantity – - the probability of finding the particle near x
• is related to the momentum probability density - - the probability of finding a particle with a particular momentum
The Wavefunction
|ψ|2 dx∣∣ dψ∣ψ∗∣ dx
∣∣∣∣ dx
7
PHYSICALLY MEANINGFUL STATES MUST HAVE THE FOLLOWING PROPERTIES:
(x) must be single-valued, and finite(finite to avoid infinite probability density)
(x) must be continuous, with finite d /dx(because d /dx is related to the momentum density)
In regions with finite potential, d /dx must be continuous (with finite d2 /dx2, to avoid infinite energies)
There is usually no significance to the overall sign of (x)(it goes away when we take the absolute square)(In fact, (x,t) is usually complex !)
ψ
x
L
(x) In what energy level is the particle? n = …
What is the approximate shape of the potential V(x) in which this particle is confined?
(a) V(x)
E
L
(b)
V(x)
E
L
(c)
V(x)
E
L
Solutions to Schrodinger s Equation
(a) 7
(b) 8
(c) 9
8
WHICH WAVEFUNCTION CORRESPONDS TO WHICH POTENTIAL WELL ?
(A)
(B)
(C)
NOTICE THAT FOR FINITE POTENTIAL WELLS WAVEFUNCTIONS ARE NOT ZERO AT THE WELL BOUNDARY 9
(1)
(2)
(3)
A Simple Potential Step
Region 1 Region 2
CASE I : Eo > V
In Region 1: In Region 2:
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = Eo
E = 0x
x = 0
2�
Eoψ = −2m
∂2ψ
∂x2
2�
(Eo − V )ψ = − ∂2ψ
2m
2k2
mEo1 =
∂x2
�2
k22m (Eo
2 =− V )
�2
V
10
A Simple Potential Step
CASE I : Eo > V
is continuous: is continuous:
ψ1 = Ae−jk1x +Bejk1x ψ2 = Ce−jk2x
ψ1(0) = ψ2(0) A+B = C
∂
∂xψ(0) =
∂
∂xψ2(0) A−B =
k2
ψ
∂ψC
k1∂x
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
11
A Simple Potential Step
CASE I : Eo > V
B
A=
1− k2/k11 + k2/k1
=k1 − k2k1 + k2
C
A=
2
1 + k2/k1
=2k1
k1 + k2
A+B = C
A−B =k2k1
C
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
ψC = Ce−jk1x
E = Eo
E = 0
x = 0x
V
12
Quantum Electron Currents
Given an electron of mass that is located in space with charge density and moving with momentum corresponding to
… then the current density for a single electron is given by
m
ρ = q |ψ(x)|2
< p > < v >= �k/m
J = ρv = q |ψ|2 (�k/m)
14
A Simple Potential Step
CASE I : Eo > V
JtransmittedTransmission = T =
Jincident=
JCJA
=|ψC |2(�k2/m)
|ψA|2(�k1/m)=
∣∣∣∣CA∣∣∣∣2k2
JreflectedReflection = R =
k1
JB=
Jincident JA=
|ψB |2(�k1/m)
|ψA|2(�k1/m)=
∣∣∣∣BA∣∣∣∣2
B 1=
− k2/k1A
C
1 + k2/k1
2=
A 1 + k2/k1
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
15
A Simple Potential Step
CASE I : Eo > V
1
T
R
T +R = 1
1
BReflection = R =
∣∣∣∣2
A
∣∣∣ =
∣∣k1 − k2∣ ∣∣∣2
k1 + k2
∣∣
Transmission = T = 1
∣−R
4k1k2= |k1 + k2|2
k2k1
=
√1− V
EoEo = V Eo = ∞
Region 1 Region 2
ψA = Ae−jk1x ψC = Ce−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
16
IBM Almaden STM of Copper
Image originally created by the IBM Corporation. © IBM Corporation. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.
17
IBM Almaden Image originally created by the IBM Corporation.
© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.
18
IBM Almaden Image originally created by the IBM Corporation.
© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse.
19
A Simple Potential Step
CASE II : Eo < V
In Region 1: In Region 2:
2�
Eoψ = − ∂2ψ
2m ∂x2
2�
(Eo − V )ψ = − ∂2ψ
2m
2k2
mEo1 =
∂x2
�2
κ2 2m (Eo=
− V )
�2
ψC = Ce−κx
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
20
A Simple
ψ2 = Ce−κx
Potential Step
is continuous: is continuous:
CASE II : Eo < V
ψ1 = Ae−jk1x +Bejk1x
ψ1(0) = ψ2(0) A+B = C
∂
∂xψ(0) =
∂
∂xψ2(0)
ψ
∂ψ κA
∂x−B = −j
ψC = Ce−κx
Ck1
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
21
A Simple Potential Step
CASE II : Eo < V
B
Total reflection Transmission must be zero
1 + jκ/k1=
A
C
1− jκ/k1
2=
A 1− jκ/k1
R =
∣∣B∣∣A∣∣∣∣2
= 1 T = 0
A+B = C
A−B = −jκ
ψC = Ce−κx
Ck1
Region 1 Region 2
ψA = Ae−jk1x
ψB = Be−jk1x
E = Eo
E = 0
x = 0x
V
22
Quantum Tunneling Through a Thin Potential Barrier
Total Reflection at Boundary
Frustrated Total Reflection (Tunneling)
R = 1 T = 0
T �= 0
23
KEY TAKEAWAYS
Region 1 Region 2
CASE II : Eo < V
A Simple Potential Step
CASE I : Eo > V
Region 1 Region 2
BReflection = R =
∣∣∣∣A∣∣∣∣2
=
∣∣∣∣k1 − k22
k1 + k2
∣∣∣
4 k
∣
Transmission = T = 1− 1k2R = |k1 + k2|2
PARTIAL REFLECTION
TOTAL REFLECTION
24
CASE II : Eo < V
Region 1 Region 2 Region 3
In Regions 1 and 3: In Region 2:
A Rectangular Potential Step
for Eo < V :
κxψA = Ae−jk1x ψC = Ce−
ψB = Bejk1x
ψF = Fe−jk1x
ψD = Deκx
VE = Eo
E = 0 −a a0
2�
Eoψ = −2m
∂2ψ
∂x2
(Eo − V )ψ = − �2
2m
∂2ψ
2k2
mEo1 =
∂x2
�2
κ2 =2m(V − Eo)
�2
T =
∣∣∣∣FA∣∣∣∣2
=1
1 + 14
V 2
Eo(V−Eo)sinh2(2κa)
25
A Rectangular Potential Step
for Eo < V :
T =
∣∣F∣∣A∣∣∣∣2
=1
1 + 14
V 2
Eo(V−Eo)sinh2(2κa)
T =
∣∣∣∣FA∣∣∣∣2
≈ 1
1 + 14
V 2
Eo(V−Eo)
e−4κa
sinh2(2κa) =[e2κa − e−2κa
]2 ≈ e−4κa
UE
26
Flash Memory
Erased 1
Stored Electrons
Programmed 0
Insulating Dielectric
CONTROL GATE Tunnel Oxide
FLOATING GATE
SOURCE CHANNEL
Substrate
Channel
Floating Gate
DRAIN
x
V (x)
Electrons tunnel preferentially when a voltage is applied
Image is in the public domain
27
MOSFET: Transistor in a Nutshell
Conducting Channel
Tunneling causes thin insulating layers to become leaky !
Image is in the public domain
Conduction electron flow
Control Gate
SemiconductorImage courtesy of J. Hoyt Group, EECS, MIT. Photo by L. Gomez
Image courtesy of J. Hoyt Group, EECS, MIT.
Photo by L. Gomez
28
Reading Flash Memory
To obtain the same channel charge, the programmed gate needs a higher control-gate voltage than the unprogrammed gate
How do we WRITE Flash Memory ? 29
CONTROL GATE
FLOATING GATE
SILICON
CONTROL GATE
FLOATING GATE
UNPROGRAMMED PROGRAMMED
0 L
V0
x
Eo
metal metal
air gap
L = 2a
Example: Barrier Tunneling
Question: What will T be if we double the width of the gap?
• Let s consider a tunneling problem:
An electron with a total energy of Eo= 6 eV approaches a potential barrier with a height of V0 = 12 eV. If the width of the barrier is L = 0.18 nm, what is the probability that the electron will tunnel through the barrier?
T =
∣∣∣∣FA∣∣∣∣2
≈ 16Eo(V − Eo)
V 2e−2κL
κ =
√2me
�2(V − Eo) = 2π
√2me
h2(V − Eo) = 2π
√6eV
1.505eV-nm2 ≈ 12.6 nm−1
T = 4e−2(12.6 nm−1)(0.18 nm) = 4(0.011) = 4.4%
30
Multiple Choice Questions
Consider a particle tunneling through a barrier:
1. Which of the following will increase the likelihood of tunneling?
a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the particle
2. What is the energy of the particles that have successfully escaped ?
a. < initial energy b. = initial energy c. > initial energy
0 L
V
x
Eo
Although the amplitude of the wave is smaller after the barrier, no energy is lost in the tunneling process
31
Application of Tunneling: Scanning Tunneling Microscopy (STM)
Due to the quantum effect of barrier penetration, the electron density of a material extends beyond its surface:
material STM tip One can exploit this to measure the ~ 1 nm electron density on a material s surface:
Sodium atoms on metal:
STM images
Single walled carbon nanotube:
V E0
STM tip material
Image originally created by IBM Corporation
© IBM Corporation. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse Image is in the public domain .
32
MIT OpenCourseWarehttp://ocw.mit.edu
6.007 Electromagnetic Energy: From Motors to LasersSpring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.